The standard electrode potentials $E^o_{(I_2/I^{-})}$,$E^o_{(Br^{-}/Br_2)}$,and $E^o_{(Fe/Fe^{2+})}$ are respectively $+0.54 \ V$,$-1.09 \ V$,and $0.44 \ V$. On the basis of the above data,which of the following processes is non-spontaneous?
- A$Br_2 + 2I^{-} \to 2Br^{-} + I_2$
- B$Fe + Br_2 \to Fe^{2+} + 2Br^{-}$
- C$Fe + I_2 \to Fe^{2+} + 2I^{-}$
- D$I_2 + 2Br^{-} \to 2I^{-} + Br_2$
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Similar Questions
Which of the following statements is not correct $:-$
Medium
View SolutionEfficiency of the following cell is $84\%$. $A_{(s)} + B^{2+}(aq.) \rightleftharpoons A^{2+}(aq.) + B_{(s)}$; $\Delta H^o = -285 \ kJ$. Then the standard $EMF$ of the cell will be ........... $V$.
Medium
View SolutionMatch the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+}|Cu$ $(II)$ $0.059$ $(C)$ $Zn|Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+}|Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn|Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
Based on the following information,arrange four metals,$A$,$B$,$C$,and $D$,in order of increasing ability to act as reducing agents:
$(I)$ Only $C$ reacts with $1 \ M \ HCl$ to give $H_{2(g)}$.
$(II)$ When $A$ is added to a solution of the other metal salts,metallic $D$ is formed,but not $B$ or $C$.
$(I)$ Only $C$ reacts with $1 \ M \ HCl$ to give $H_{2(g)}$.
$(II)$ When $A$ is added to a solution of the other metal salts,metallic $D$ is formed,but not $B$ or $C$.
The rusting of iron takes place as follows. Calculate $\Delta G^o$ for the net process in $kJ \ mol^{-1}$.
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$
$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$
$2H^{+} + 2e^- + \frac{1}{2}O_2 \longrightarrow H_2O_{(l)} ; E^o = +1.23 \ V$
$Fe^{2+} + 2e^- \longrightarrow Fe_{(s)} ; E^o = -0.44 \ V$
DifficultIIT 2005
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